If I compare with the last section then this seems to be what one should do. Because In that case

$$x\rightarrow \lambda\ x,$$
$$t\rightarrow \lambda\ t,$$
$$\rho\rightarrow \lambda\ \rho,$$
$$u\rightarrow u,$$
and therefore the naviar stokes equation with
$$...+\rho\ u\ \nabla u+\mu\ \nabla^2 u+...$$
is such, that its invariant if the derivative produces an additional $\lambda$, the way I did above.

Sidenote: is scale invariance $\phi(x)=\phi^{-\Delta}(\lambda x)$ of fields something that should more acurately be called self similarity?

Also, I don't understand the last sentence in the $\phi^4$ paragraph, which says that the dimensionlessness of the coupling $g$ is "the main point". Why is it a problem to also scale the coupling? However, I never understood why dimensionless couplings are prefered anyway. If the coupling is somehow unique, then why does it make sense to search for a process dependend coupling in renormalization theory?

1 Answer
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The reason you're not finding scale invariance in your equation is that you've inserted a factor of $\lambda^2$ in the first two terms, which as far as I can see has no reason to be there.

I'm guessing you put it in because of your uncertainty as to how to dilate derivatives. Transforming a derivative is nothing complicated, though. You simply make the transformation on the variable that is being differentiated with respect to, treat it as a differential in the denominator, and simplify. For example:

When you do this, in each of the first two terms you get $\lambda^{-1}$ from the transformation of the field and $\lambda^{-2}$ from the transformation of the derivative, for an overall scaling factor of $\lambda^{-3}$. And in the last term, you get $\lambda^{-3}$ from the fields alone. All those factors of $\lambda^{-3}$ then cancel out.

The reason dimensionless couplings are preferred (in certain applications) is that they don't set any scale for the theory. If your coupling $g$ had a dimension of $L^{-1}$, for example, then you could establish a characteristic length scale for that theory as $g^{-1}$. But being scale invariant means that there should be no preferred length scale; you can change any scale $x \to \lambda x$ without altering the theory. If you have a dimensionful coupling, that is obviously not the case, because you can distinguish between $x$ and $\lambda x$ by measuring them in units of $g^{-1}$. You can get around that by having $g$ change under a scale transformation as well, but then it's no longer a coupling constant, which makes things more complicated.

I thought of the product rule when substituting $\frac{∂}{∂x}$ with $\lambda\frac{∂}{∂x}$. Of course I thought about doing it like you did, but as I pointed out in the question, I don't see how that procedure fits with the other scaling example. If I consider these two terms, then it seems like $\nabla$ should scale like $\rho$. Also what exactly does "which makes things more complicated" mean? Does it mean that one practically does this? And why is $g^{-1}$ the length scale, doesn't that depend on many things?
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NikolajKJan 29 '12 at 11:21

No need for the product rule here. I think the dynamic viscosity has to scale as $\mu\to\lambda^{2}\mu$, otherwise something is wrong with the argument about scale invariance of the Navier-Stokes equations. Changing the coupling constant under a scale transformation gets into aspects of renormalization which I don't understand as well as I should. And what are these many things you're saying the length scale depends on? If $g$ has units of $L^{-1}$, then $g^{-1}$ is a length, simple as that.
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David Z♦Jan 29 '12 at 21:13

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@Nick Kidman and David Zaslavsky: There was an error on wikipedia in the scaling law for $\rho$, which is now corrected.
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Qmechanic♦Jan 29 '12 at 23:20